If a 6.0keV photon scatters from a fee proton at rest, what is the change in the photon's wavelength if the photon recoils at 90 degrees?
To find the change in the photon's wavelength when it scatters from a free proton at rest, we can use the Compton scattering formula:
Δλ = λ' - λ = λ * (1 - cosθ),
where Δλ is the change in the wavelength, λ' is the final wavelength, λ is the initial wavelength, and θ is the scattering angle.
We are given that the initial photon has an energy of 6.0 keV. We need to convert this energy into wavelength using the equation:
E = hc/λ,
where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength.
Rearranging the equation, we get:
λ = hc/E.
Substituting the values, we have:
λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (6.0 x 10^3 eV * 1.602 x 10^-19 J/eV).
Calculating this, we find that λ ≈ 2.476 x 10^-12 m.
Next, we know that the scattering angle is 90 degrees (θ = 90°).
Using this information, we can now find the change in wavelength:
Δλ = λ * (1 - cosθ).
Plugging in the values, we have:
Δλ = (2.476 x 10^-12 m) * (1 - cos(90°)).
cos(90°) = 0, thus:
Δλ = (2.476 x 10^-12 m) * (1-0).
Simplifying further, we get:
Δλ = (2.476 x 10^-12 m) * 1.
Finally, Δλ ≈ 2.476 x 10^-12 m.
Therefore, the change in the photon's wavelength when it scatters from a free proton at rest and recoils at 90 degrees is approximately 2.476 x 10^-12 meters.